Chapter 4.3, Problem 41AYU

To determine

To analyze: the graph of given function

Answer to Problem 41AYU

  

Explanation of Solution

Given:

  $f\left(x\right)=x^2+\frac{1}{x}$

Calculation:

STEP 1: Rewrite the function as $f\left(x\right)=\frac{x^3+1}{x}$

Now, factor the function $f$ .

  $f\left(x\right)=\frac{\left(x+1\right)\left(x^2-x+1\right)}{x}$

The domain of $f$ is $\{x|x\cancel{=}0\}$ .

Since $0$ is in the domain, there is no $y-intercept$ .

STEP 2: The function $f$ is in lowest terms. Since $x+1=0$ has the real solution $x=-1,-1isthex-intercept$ .

Determine the behavior of the graph near the $x-intercept$ .

  $\begin{array}{l}R\left(x\right)=\frac{x^3+1}{-1}\\ \approx -1\left(x^3+1\right)\end{array}$

Plot the point $\left(-1,0\right)$ and indicate a line with slope $-1$ on the graph.

STEP 3: $f$ is in lowest terms. So, $x=0$ is a vertical asymptote of $f$ . Graph the line $x=0$ using a dashed line.

STEP 4: Degree of the numerator is greater than the degree of the denominator. So, the rational function is improper. To find any horizontal or oblique asymptote, we use the long division.

  $x\begin{array}{c}\hfill x^2\\ \hfill \overline{)\begin{array}{l}{x}^3+1\\ \frac{x^3}{\text{ 1}}\end{array}}\end{array}$

The quotient is $x^2$ . So, the graph does not have any horizontal or oblique asymptote. But the graph of $f$ will approach the graph of $y=x^2\text{ as }x\to -\infty andx\to \infty$ . The graph of $f$ does not intersect with $y=x^2$ .

STEP 5: The numerator has one zero at $-1$ and the denominator has one zero at $0$ . Therefore, we divide the $x-axis$ into three intervals. The three intervals are $\left(-\infty ,-1\right),\left(-1,0\right),\left(0,\infty \right)$ .

Now, construct the table.

  

Interval Number chosen Value of f Location of graph Point on graph

$\begin{array}{l}\left(-\infty ,-1\right)\\ -2\\ f\left(-2\right)=3.5\\ Abovex-axis\\ \left(-2,3.5\right)\end{array}$

$\begin{array}{l}\left(-1,0\right)\\ -\frac{1}{2}\\ f\left(-\frac{1}{2}\right)=-1.75\\ Belowx-axis\\ \left(-\frac{1}{2},-1.75\right)\end{array}$

$\begin{array}{l}\left(0,\infty \right)\\ 1\\ f\left(1\right)=2\\ Abovex-axis\\ \left(1,2\right)\end{array}$

Plot the points from the table obtained.

  

STEP 6: Since the graph of $f$ is above the $x-axisforx<-1$ , we place arrows above the $x-axis$ as shown in the figure. Similarly, as the graph of $f$ is below the $x-axisfor-1<x<0$ , we place arrows below the $x-axis$ . Also, as the graph of $f$ is above the $x-axisforx>0$ , we place arrows above the $x-axis$ as shown in the figure.

  

STEP 7: The figure shows the final graph.

  

Conclusion:

Thus, the graph of $f$ is obtained which illustrates the information obtained from Steps 1 through 6.